Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{t^2 - 4t}{t^2 + t - 20} \div \dfrac{2t^2 + 14t}{t^2 + 4t - 21} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{t^2 - 4t}{t^2 + t - 20} \times \dfrac{t^2 + 4t - 21}{2t^2 + 14t} $ First factor out any common factors. $y = \dfrac{t(t - 4)}{t^2 + t - 20} \times \dfrac{t^2 + 4t - 21}{2t(t + 7)} $ Then factor the quadratic expressions. $y = \dfrac {t(t - 4)} {(t - 4)(t + 5)} \times \dfrac {(t + 7)(t - 3)} {2t(t + 7)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {t(t - 4) \times (t + 7)(t - 3) } { (t - 4)(t + 5) \times 2t(t + 7)} $ $y = \dfrac {t(t + 7)(t - 3)(t - 4)} {2t(t - 4)(t + 5)(t + 7)} $ Notice that $(t - 4)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {t(t + 7)(t - 3)\cancel{(t - 4)}} {2t\cancel{(t - 4)}(t + 5)(t + 7)} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $y = \dfrac {t\cancel{(t + 7)}(t - 3)\cancel{(t - 4)}} {2t\cancel{(t - 4)}(t + 5)\cancel{(t + 7)}} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $y = \dfrac {t(t - 3)} {2t(t + 5)} $ $ y = \dfrac{t - 3}{2(t + 5)}; t \neq 4; t \neq -7 $